The η-meter Part IV says every multiplet enters through three integers. This page turns that into an instrument: it predicts what the boundary sign η does to the Higgs mass matrix without evaluating the potential at all, and then evaluates it to show you the prediction was right. Below, the same potential with the field released on it.

verified — Part IV closed form theorem · Lean 4 — blindness measured — prediction vs brute force self-check: running…

0 · What this is about, in one screen i

the field
The Higgs is a Wilson line. In gauge–Higgs unification the extra components A₅, A₆ of a six-dimensional gauge field are the Higgs doublet; its vacuum value is the pair of phases α = (α₁, α₂) a Wilson line picks up around the two cycles of the torus. So “where the Higgs sits” is literally a point on a torus, and the Higgs potential is a function on that torus.
the potential
There is no Higgs potential at tree level — gauge invariance forbids it. It appears at one loop, as a sum over the windings k of the torus of every mode the matter content carries: V(α) = Σk≠0|k|⁻⁶ Σq[Aq + Bq(−1)k₂]cos(2πq·kα). The Higgs mass is the curvature of that at its minimum. That is why the mass is calculable here and not a free parameter.
the boundary sign
On the orbifold T²/Z₂ each multiplet carries a discrete choice η = η₀η₁ = ±1 — how it comes back to itself around the fixed points. It multiplies the coset half of the spectrum and nothing else, so it is a genuinely physical knob: flip it and the Higgs mass can change.
what this page adds
The coset half is an index Dλ = A − B, so it can cancel. Part IV says D is a product of three SU(2) characters — three integers per multiplet. This page uses that to answer, by arithmetic and in no time at all, what η does to the Higgs for any content you type, and then does the honest thing: it evaluates the loop sum and shows you both numbers.

1 · Your matter content i

(a,b,c)dimbox (p,q,r)M₀ η
multipletdimn ηrole
Pick multiplets above, or load AHMN's content.

2 · The answer i

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The five formulas behind that answer — open only if you want the machinery
F1 · The box — a whole multiplet, in three integers verified
Dλ(j) = ζ · (χp ∗ χq ∗ χr)(j) (p,q,r) read off the 2-quotient of λ
goes in
One irreducible multiplet, as a partition λ of four rows — of any size.
comes out
Three integers and a sign. The coset index D is the convolution of three SU(2) characters, and nothing else about λ survives.
what it buys you
The η-dependence of a multiplet without enumerating its Kaluza–Klein tower. A 3000-dimensional representation costs the same as a singlet.
F2 · The moments — three numbers you are allowed to add up verified
M₀ = ζ (p+1)(q+1)(r+1) Ck = k(k+2), e₁ = Cp+Cq+Cr M₂ / M₀ = e₁ / 3 M₄ / M₀ = (2/3)e₂ + (1/5)(C²p+C²q+C²r) − (4/15)e₁
goes in
The box of each multiplet in your content, with its multiplicity n and its boundary sign η.
comes out
The moments of the coset index. They are linear in the content: M(total) = Σ n·η·M(λ), so a model is the sum of its parts here.
what it buys you
A model of any size is compressed to three numbers — and M₂ is the one the physics below turns out to need.
F3 · The inversion — run the compression backwards verified
e₁ = 3M₂/M₀ e₂ = (15/4)(M₄/M₀) − (3/4)e₁² + e₁ e₃ = M₀² − 1 − e₁ − e₂ {Cp, Cq, Cr} = roots of x³ − e₁x² + e₂x − e₃ k = √(C+1) − 1
goes in
Three moments — M₀, M₂, M₄ — of anything, a single multiplet or a whole model.
comes out
The box back, exactly. Three moments suffice; M₀ is not a normalisation, it carries e₃.
what it buys you
You can ask a model whether it behaves like a single multiplet. If the inversion lands on integers, your whole content has an effective box; if it does not, it is a genuine mixture. That is not visible by looking at the content.
F4 · The prediction — the whole effect of η, in one number derived here · checked against brute force
ΔHij ≡ Hij(η) − Hij(−η) at α = 0 ΔH11 = −2(2π)² L₁ · S₂ ΔH22 = −2(2π)² L₂ · S₂ ΔH12 = 0 L₁ = Σk₂ odd |k|⁻⁶ k₁² L₂ = Σk₂ odd |k|⁻⁶ k₂² (the lattice, once and for all) S₂ = M₂/8
goes in
M₂ of your content — one number, from F2, with no sum over windings and no potential evaluated.
comes out
The complete change of the Higgs mass matrix at the symmetric point when the boundary sign flips. The off-diagonal is exactly zero, by the parity of the lattice sum.
what it buys you
Scan models by arithmetic instead of by numerics: any content with M₂ = 0 is invisible to η at this order however large it is, and the sign of M₂ says which way the boundary sign pushes the Higgs.
F6 · The search — can this content be made blind on purpose? exact integer arithmetic
Dtotal(j) = Σλ nλ ηλ Dλ(j) ≡ 0 with nλ ∈ {1…6}, ηλ = ±1
goes in
The multiplets you have chosen — the sighted ones; blind ones are along for free.
comes out
Multiplicities and boundary signs for which the coset index cancels identically, charge by charge, in exact integers — or the honest statement that none does.
what it buys you
Collective blindness. A model built entirely from multiplets that each see η, and which as a whole cannot: η becomes unobservable in this Higgs sector not by a theorem about one multiplet but by your choice of content.
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F5 · Blindness — when η cannot be measured at all theorem · Lean 4
Dλ ≡ 0 ⟺ λ₁ ≢ λ₂ ≢ λ₃ ≢ λ₄ (alternating parity) or λi + λ5−i = c for all i, with c odd (self-complementary, odd constant)
goes in
The partition alone — no character, no spectrum, no potential.
comes out
Whether the boundary condition has any measurable consequence for that multiplet.
what it buys you
A model built only from blind multiplets cannot be told apart from its η-flipped twin by anything in this Higgs sector. F4 is the graded version of the same statement: blind ⇒ M₂ = 0.

3 · Let the field roll i

The Wilson line is the Higgs here, so the potential above is what the field falls down. Release many configurations at once and watch where they end up: the basins are what decides which vacuum a region of the universe picks. This is the zero mode relaxing — one α for all of space, with friction — not a lattice field simulation, and it is drawn on the same V the rest of the page computes.

plan · where they land
relief · the same V

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4 · The atlas — every multiplet's landscape, side by side i

One thumbnail per multiplet: the one-loop potential that multiplet alone would make, over the same square, on the same colour scale as itself. This is the page as a research instrument: put a hundred landscapes next to each other and the structure Part IV predicts is visible without reading a number — multiplets that share a box share a landscape, and the corrugation grows with p+q+r. Switch to η-difference and every blind multiplet goes blank: that is Theorem 1, as a picture, a hundred times at once.

What is what. verified F1–F3 are Part IV's closed form and its inversion: exhaustively checked, not proved. theorem F5 is Part V, machine-checked in Lean 4. measured F4 is derived here from the one-loop sum and is checked on this page against the brute-force Hessian every time you change the content — if the two columns ever disagree, the page says so instead of hiding it.

Papers. Part III 10.5281/zenodo.21438226 · Part IV 10.5281/zenodo.21463000 · Part V 10.5281/zenodo.21727094 · code higgs-blind-class. Carles Marín Muñoz, independent researcher, with Claude (Anthropic) as AI research assistant. Apache 2.0. Nothing you type leaves this page.